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Sum of Prime Numbers


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Spoiler

Consider the set of integers 2, 3 ... 99. After removal of the first two primes (2, 3) we are left with 96 members, 48 of which are divisible by 2, leaving 48, of which a further 16 are divisible by 3. The remaining 32 plus the members 2 & 3 leave a maximum of 34 potential primes less than 100. Therefore, at least 134 primes <1000 are greater than 100. The sum of these must exceed 13,400 therefore option (A) is false.

Of the 168 total, 167 of these are odd numbers and sum to an odd number. Therefore options (C) & (D) are both false.

Only option (B) survives. 

 

Edited by sethoflagos
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You are correct, +1.

Would you please use the "Spoiler" feature next time, to give others a chance?

 

Spoiler

The first part can be made shorter: the sum of first 168 integers is 168*169/2 = 14196. This immediately eliminates option A.

 

Edited by Genady
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Just now, sethoflagos said:

Apologies. Not aware there was one. Where do I find it?

It is the "eye" icon above on the very right.

4 minutes ago, sethoflagos said:

Done.

Thank you.

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6 hours ago, Genady said:

I've stumbled upon this puzzle and immediately thought of @Trurl  because they post about prime numbers often. Of course, others are welcome to try it as well.

image.png.ea71f5360208cf2a7437a921c82e17ff.png

Spoiler

As you add primes, you alternate between odd and even sums...5,10,17,28,41...and so on.   So you will end the addition process on an odd number, so C and D are out.  And A is too small, even just summing the first 168 integers will yield a larger number.  So B.

Thanks, that was fun.

 

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17 minutes ago, TheVat said:
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As you add primes, you alternate between odd and even sums...5,10,17,28,41...and so on.   So you will end the addition process on an odd number, so C and D are out.  And A is too small, even just summing the first 168 integers will yield a larger number.  So B.

Thanks, that was fun.

 

Correct, and glad you liked it. +1

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